Notation. When I say "ring", I mean "ring with unity" (not necessarily commutative).

Definition. A ring $R$ is said to be left-Artinian if for every sequence $I_0\supseteq I_1\supseteq I_2\supseteq I_3\supseteq ...$ of left ideals of $R$, there exists an $n\in\mathbb N$ such that $I_n=I_{n+1}$.

Definition. A ring $R$ is said to be right-Artinian if for every sequence $I_0\supseteq I_1\supseteq I_2\supseteq I_3\supseteq ...$ of right ideals of $R$, there exists an $n\in\mathbb N$ such that $I_n=I_{n+1}$.

Definition. A ring $R$ is said to be Artinian if it is both left-Artinian and right-Artinian.

Definition. The Jacobson radical $\operatorname{Ra}R$ of a ring $R$ is defined by one of the following equivalent definitions:

(where "f.g." means "finitely generated"). (Note that the equivalences are constructive; I have written up the proofs in German a while ago (search for "Jacobson-Radikal") and will translate when I have the time.)

Definition. A ring $R$ is said to be von Neumann regular if for every $r\in R$, there exists some $x\in R$ such that $rxr=r$.

Question: Can we constructively prove that every Artinian ring $R$ satisfying $\operatorname{Ra}R=0$ is von Neumann regular? (This is proven classically using the AC in Lam, "A first course in noncommutative rings", Theorem (4.14) + Corollary (4.24).)

Normally, theorems in algebra can be proved constructively if we know a classical proof. There are methods for this (scindage a la Lombardi; dynamic proofs; Gödel-Gentzen etc.). Unfortunately, whenever chain conditions (such as Artinianity) are involved, these methods break down. The constructive Artinian condition is neither easy to use nor easy to satisfy, so I am not completely sure whether the question is the right one to ask - but I don't know of a better one.

While constructive Artinianity is far less useful than classical Artinianity, it can still be applied to chains of ideals such as $R\supseteq rR\supseteq r^2R\supseteq r^3R\supseteq ...$ to conclude that for every $r\in R$ there exists some $n\in\mathbb N$ and some $y\in R$ such that $r^n=r^{n+1}y$. This can then be juggled with (for example, we can conclude that $r^n=r^ayr^b$ for any two nonnegative integers $a$ and $b$ with $a+b=n+1$; here we use $\operatorname{Ra}R=0$). This is, at the moment, my main reason to believe that the Question above has a positive answer (we mainly have to bring the $n$ down to $1$). But, as I said, I am far from sure about this.

Meta-question: What is the (morally) right constructive analogue of the notions "Artinian" and "Noetherian"? Given the definition of "Artinian" above, I am not sure if $\mathbb F_2$ is Artinian, because I could take a sequence $S_0,S_1,S_2,...$ of independent statements which are independent of each other too (is this possible?) and then let $I_n$ be the ideal containing $1$ if $S_0, S_1, ..., S_{n-1}$ hold. So let me pose a different question, which is actually interesting without relying on constructivity:

Concrete question. Let $R$ be a ring with $\operatorname{Ra} R = 0$. Assume that, for every $r \in R$, there exists an $n \in \mathbb{N}$ and a $y \in R$ such that $r^n = r^{n+1} y$. Also assume that, for every $r \in R$, there exists an $n \in \mathbb{N}$ and a $y \in R$ such that $r^n = y r^{n+1}$. Must $R$ then be von Neumann regular?

Notice that this is NOT a constructive translation of the classical theorem. It is a stronger conjecture which has the advantage of not requiring a constructive translation to begin with.

Not to be nitpicky, but (here begins nitpickiness) in your definition of left and right Artinian, you need $I_n = I_{n+1}$ for all $n \ge M$ for some $M$.
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MTSMar 8 '11 at 21:26

I'm just curious, by "classical artinianity" do you mean the condition that every set of left ideals has a minimal element?
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Manny ReyesMar 8 '11 at 21:46

By "classical Artinianity" I mean what MTS has written. Constructively it is a FAR too strong condition (not even the field with two elements is constructively classically Artinian, because this would solve the halting problem).
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darij grinbergMar 8 '11 at 21:54

So you really want to work with the above definition of artinian, which is not the usual one at first sight? By the way I've deleted the tag "von Neumann algebras" since they are special $C^*$-algebras completely unrelated to von Neumann regular rings.
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Martin BrandenburgMar 8 '11 at 23:08

Yes, I want to work with my definition of Artinian - unless somebody gives me a better one (which I hope). The usual one is utterly useless in constructive mathematics.
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darij grinbergMar 8 '11 at 23:15

2 Answers
2

The answer to the new concrete question is no. Your question boils down to whether there exists a semi-primitive strongly $\pi$-regular ring which is not von Neumann regular. The answer is yes. For instance, you can use the subring of $\prod_{\omega}\mathbb{M}_2(\mathbb{F}_2)$ consisting of sequences of matrices which are eventually stable (meaning the matrices stop changing after some point) and eventually upper-triangular.

By the way, suppose we revert back to the original question of whether it is possible to find a constructive proof that left/right artinian and semiprimitivity implies v.N. regularity, and we also assume that semiprimitivity gives us an oracle such that for every element $r\in R$ it gives us $s\in S$ such that $1-rs$ is not invertible. Using a construction similar to this one, I believe I can show that we cannot prove v.N. regularity (essentially because of the example above).

Very nice counterexample! So your subring essentially steals its zero-radical property from the $M_2\left(\mathbb F_2\right)$ factors, its lack of von Neumann regularity from the (limit) ring of upper-triangular matrices, and its strong $\pi$-regularity from both.
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darij grinbergJan 26 at 15:53

I'm not as familiar with constructive proofs, so this "answer" is really just a couple questions.

Consider the ring $R=\mathbb{M}_2(\mathbb{F}_2)$, and the element $r=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}$. Let's treat $r$ as the input in our algorithm. Without appealing to the semiprimitivity or artinian conditions, the only constructable elements are integer polynomials in $r$. (Is this correct?) Thus, we have access to $0,1,r,1+r$, and nothing else (since $2=0$ and $r^2=0$; although we do know these two equalities).

The left ideals generated by these elements are $(0)\subsetneq Rr\subsetneq R$, and the right ideals are $(0)\subsetneq rR\subsetneq R$ (and these containments are provable with what we constructively know). Thus, appealing the the left and right artinian conditions will not yield anything.

The element $r$ is nilpotent. Since the Jacobson radical is zero, we know that some multiple of $r$ must not be nilpotent. But do we have access to any constructible element $s$ such that $rs$ is not nilpotent? What kind of elements does a zero Jacobson radical hypothesis allow us claim existence for?

I'm not really sure about this all. But I don't think that you can constructively prove that the Jacobson radical is zero for a ring which "floats" somewhere between $\left\{0,1,r,1+r\right\}$ and the whole $R$ (e.g., the ring containing $0,1,r,1+r$ and also containing $r^T$ if some given undecidable statement is true). But Artinianity is also far from obvious for these rings. Let me edit my OP actually.
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darij grinbergJan 26 at 13:06